$g(n) = -n^{3}-4n^{2}$ $h(t) = 7t^{2}+6t+3(g(t))$ $ h(g(-1)) = {?} $
Answer: First, let's solve for the value of the inner function, $g(-1)$ . Then we'll know what to plug into the outer function. $g(-1) = -(-1)^{3}-4(-1)^{2}$ $g(-1) = -3$ Now we know that $g(-1) = -3$ . Let's solve for $h(g(-1))$ , which is $h(-3)$ $h(-3) = 7(-3)^{2}+(6)(-3)+3(g(-3))$ To solve for the value of $h$ , we need to solve for the value of $g(-3)$ $g(-3) = -(-3)^{3}-4(-3)^{2}$ $g(-3) = -9$ That means $h(-3) = 7(-3)^{2}+(6)(-3)+(3)(-9)$ $h(-3) = 18$